Re: Why? [was Re: Cantor`s powerset theorem is false?]

apoorv wrote:



A while back, after having already corrected a number of your posts, I
said I'd read and correct two more. And now I am finishing that
promise. And bravo to you for starting your post, again, IMMEDIATELY
with a GLARING error of predicate logic and display of misunderstanding
of the axioms of set theory. You're like a jack-in-the box who keeps
popping up to hit me over the head with the hammer of your OWN
fallacies. Each time, you project YOUR fallacies into MY proofs. Either
you are just playing games doing that or you really do not understand
the most basic logic (even failing to understand the most basic logic
of conditional proof, as I indulged you with an explanation in my last
post to you).

You started with Assume,S = {n in w | n not in n},
or, S ={n:n e w } intersection {n : n not in n }.

That's your very first sentence. And it's DRAMATICALLY incorrect. You
tried to INJECT YOUR fallacy into MY proof. I did NOT assume:

S = {n|new}/\{n|~nen}.

I used:

S = {new|~nen}.

YOU made the fallacious inference to:

{new|~nen} = {n|new}/\{n|~nen}.

The axioms of Z set theory do NOT allow that inference.

I'll explain in gruesome detail:

I get to infer S = {new|~nen} (with 'S' a temporary variable of
existential instantiation) as follows:

AzExAn(nex <-> (nez & ~nen)) instance of axiom schema of separation

ExAn(nex <-> (new & ~nen)) by universal instantiation

An(neS <-> (new & ~nen)) by existential instantiation

neS <-> (new & ~nen) by universal instantiation

S = {n|new & ~nen} by definition of set abstraction notation applied to
a PROVEN existing set

S = {new|~nen} by notational convention for set abstraction

Now, the REASON you can NOT infer

S = {n|new}/\{n|~nen}

is that {n|~nen} is NOT proven to exist. In general, the axiom schema
of separation:

any generalization of the closure of (i.e. universal quantifiers at the
front to bind all free variables in phi)
AzExAn(nex <-> (nez & phi)
where x does not occur free in phi

That gives the existence of

{n|nez & phi}

but it does NOT give the existence of

{n|nez}/\{n|phi}.

The reason is that to form an intersection on sets represented by
abstraction notation, you can only do so with sets PROVEN to exist. And
{n|phi} is NOT proven to exist in this case where phi is ~nen.

One would naively THINK that {n|nez & phi} = {n|nez}/\{n|phi}, because
they look so similar. And we do sometimes INFORMALLY talk about {n|nez
& phi} as z/\{n|phi} where {n|phi} may be a PROPER CLASS. But we back
off REAL FAST from making any illicit inferences from such informal
renderings, since they will IMMEDIATELY allow a contradiction that is
NOT allowed by the axioms of Z nor ZF (there are OTHER theories that
allow for the existence of proper classes, but those theories have
their own protections from inconsistency and are not to be conflated
with Z or ZF).

The axiom schema of separation allows the inference of the existence of
any set such as {nez|phi}, but we are NOT in general allowed (there is
no axiom that will in general justify the inference) that that set is
an intersection of the form {n|nez}/\{n|phi}. We can't make that
inference UNLESS we have PROVEN that {n|phi} exists, which, in the case
of {n|~nen}, we have NOT proven it to exist; indeed we have proven it
NOT to exist.

There you have it: My argument does NOT entail, as you fallaciously
argue, the existence of a universal set. And you've used up any last
credibility that you are not either a joker or simply in bad faith to
be arguing about set theory when you have not even taken the least bit
effort to make yourself competent in predicate logic (even in
sentential logic as basic as conditional derivation) or with the axioms
of Z set theory.

I hope I can keep my promise to myself now to respond with just the
word 'WRONG' to any subsequent fallaciousness in your posts and let you
figure out for yourself, if you're not a joker and don't already know,
the nature of your mistakes.

MoeBlee

.

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